poj2010 Moo University

思路：

先按照分数排序，并用大顶堆预处理出每个位置i的左侧最小N/2个花费的和minL[i]以及右侧最小N/2个花费的和minR[i]。

之后在合法范围内从大到小枚举分数中位数，对于某个位置i如果minL[i] + minR[i] + a[i].f <= F，则可行。

实现：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;

ll N, C, F, minL[100005], minR[100005];
struct node
{
    ll c, f;
};
node a[100005];

bool cmp(const node & a, const node & b)
{
    return a.c < b.c;
}

bool check(int i)
{
    return minL[i] + minR[i] + a[i].f <= F;
}

ll solve()
{
    ll s = 0;
    priority_queue<ll> q;
    for (int i = 0; i < C; i++)
    {
        if (i < N / 2)
        {
            q.push(a[i].f);
            s += a[i].f;
        }
        else
        {
            minL[i] = s;
            if (a[i].f <= q.top())
            {
                s -= q.top();
                q.pop();
                q.push(a[i].f);
                s += a[i].f;
            }
        }
    }
    s = 0;
    while (!q.empty())
    {
        q.pop();
    }
    for (int i = C - 1; i >= 0; i--)
    {
        if (i > C - 1 - N / 2)
        {
            q.push(a[i].f);
            s += a[i].f;
        }
        else
        {
            minR[i] = s;
            if (a[i].f <= q.top())
            {
                s -= q.top();
                q.pop();
                q.push(a[i].f);
                s += a[i].f;
            }
        }
    }
    for (int i = C - 1 - N / 2; i >= N / 2; i--)
    {
        if (check(i))
        {
            return a[i].c;
        }
    }
    return -1;
}

int main()
{
    while (cin >> N >> C >> F)
    {
        for (int i = 0; i < C; i++)
        {
            scanf("%I64d %I64d", &a[i].c, &a[i].f);
        }
        sort(a, a + C, cmp);
        cout << solve() << endl;
    }
    return 0;
}